hdu 3507 斜率优化dp入门

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#队列 #斜率优化 #dp

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本文介绍了一种算法问题，即如何以最小的成本打印一篇文章。该问题考虑了每打印一个单词的成本以及一行中打印多个单词的固定成本。通过动态规划方法求解最优方案，并提供了完整的代码实现。

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Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6389 Accepted Submission(s): 1965

Problem Description

Zero has an old printer that doesn't work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

Output

A single number, meaning the mininum cost to print the article.

Sample Input

Sample Output

Author

Xnozero

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

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参考blog ：http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>


using namespace std;

#define maxn 500005

typedef long long ll;

ll sum[maxn];
ll dp[maxn];
int q[maxn];
int head=0, tail=-1;
int n, m;

ll sq(ll tmp)
{
    return tmp*tmp;
}

int size()
{
    return tail-head+1;
}



ll getk_up(int i, int j)
{
    return dp[i]+sq(sum[i])-dp[j]-sq(sum[j]);
}

ll getk_down(int i, int j)
{
    return (sum[i]-sum[j])*2;
}


int main()
{

    while(scanf("%d%d", &n, &m)!=EOF){
        sum[0] = 0;
        for(int i = 1; i <= n; i++){
            scanf("%I64d", sum+i);
            sum[i] += sum[i-1];
        }


        head = 0, tail = -1;
        dp[0] = 0;
        q[++tail] = 0;
        for(int i = 1; i <= n; i++){
            while(size()>= 2 && getk_up(q[head+1],q[head])<=sum[i]*getk_down(q[head+1],q[head]))
                head++;

            int pos = q[head];
            dp[i] = dp[pos]+sq(sum[i]-sum[pos])+m;

            while(size()>=2 && getk_up(i,q[tail])*getk_down(q[tail],q[tail-1])<=getk_up(q[tail],q[tail-1])*getk_down(i, q[tail]))
                tail--;

            q[++tail] = i;
        }

        printf("%I64d\n", dp[n]);
    }
    return 0;
}